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University of Palestine Faculty of Information Technology ITGD4103 Data Communications and Networks Lecture-9: Communication Techniques ,Spectrum and bandwidth week 10- q-2/ 2008 Dr. Anwar Mousa 3-2008 UP-Copyrights reserved 1 Communication Techniques 3-2008 UP-Copyrights reserved 2 Major topics 3-2008 This chapter explains the differences between analog and digital transmission discusses Transmission impairments and channel capacity describes how digital data can be encoded by means of a modem so that they can be transmitted over analog telephone lines UP-Copyrights reserved 3 Analog and digital Analog data vs. digital data Analog signal vs. digital signal 3-2008 Analog: Continuous values on some interval (sound, light, temperature, pressure) Digital: Discrete values (text, integers, binary) Analog: Continuously varying electromagnetic wave Digital: Series of voltage pulses (square wave) UP-Copyrights reserved 4 Analog Data-->Signal Options Analog data to analog signal Inexpensive, easy conversion (e.g. telephone) Data may be shifted to a different part of the available spectrum (multiplexing) Used in traditional analog telephony Analog data to digital signal (PCM) Requires a codec (encoder/decoder) 3-2008 sender converts the voice data by a bit stream, receiver reconstruct the bit stream to the analog data Allows use of digital telephony UP-Copyrights reserved 5 Digital Data-->Signal Options Digital data to analog signal Digital data to digital signal (LINE CODING) 3-2008 Requires modem (modulator/demodulator) Allows use of PSTN (public-switched Telephone Network) to send data Necessary when analog transmission is used More reliable because no conversion is involved Less expensive when large amounts of data are involved UP-Copyrights reserved 6 3-2008 UP-Copyrights reserved 7 Advantages of digital transmission (1/2) Cost Data integrity 3-2008 the LSI and VLSI technologies has caused a continuing drop in the cost and size of digital circuitry the maintenance costs for digital circuitry are a fraction of those for analog circuitry with the use of digital repeaters, the effects of noise and other signal impairments are not cumulative it is possible to transmit data longer distances and over lesser-quality lines while maintaining the integrity of the data UP-Copyrights reserved 8 Advantages of digital transmission (2/2) Capacity utilization it has become economical to build transmission links of very high bandwidth it is more easily and cheaply achieved with digital transmission for a high degree of multiplexing to effectively utilize the capacity Security and privacy 3-2008 including satellite channels and optical fiber encryption techniques can be readily applied to digital data and to analog data that have been digitized UP-Copyrights reserved 9 Digital signals 3-2008 UP-Copyrights reserved 10 Digital signals Digital signals usually refers to the transmission of electromagnetic pulses 3-2008 that represent two binary digits, 1 and 0 binary information is generated and then converted into digital voltage pulses for transmission UP-Copyrights reserved 11 3-2008 UP-Copyrights reserved 12 EXAMPLES 3-2008 UP-Copyrights reserved 13 1. Given the frequencies listed below, calculate the corresponding periods. a. 24 Hz b. 8 MHz c.140 KHz Solution a. F = 24 Hz T= 1/f T= 1/24 = 0.0416= 41.6 ms b. F = 8 MHz T= 1/f T= 1/(8 x 106)= 0.125µs c. F = 140kHz T= 1/f T= 1/(140 x 1000)= 7.14 µs 3-2008 UP-Copyrights reserved 14 2. Given the following periods, calculate the corresponding frequencies. • a. 5 s b. 12 μs c. 220 ns Solution a. f=1/T = 1/5= 0.2 Hz b. f=1/T = 1/(12 x 10-6) = 0.083 x 106 Hz c. f=1/T = 1/(220 x 10-9) = 4.5 MHz 3-2008 UP-Copyrights reserved 15 4. What is the bandwidth of a signal that can be decomposed into five sine waves with frequencies at 20,50,100, and 200 Hz? All peak amplitudes are the same. Draw the bandwidth. Solution B = fh - fl = 200 - 20 = 180 Hz 20 50 100 200 3-2008 200 20 180 UP-Copyrights reserved 16 5. What is the transmission time of a message sent by a station if the length of the message one million bytes (each byte consists of 8 bits) and the data transmission rate is 200 kbps? Solution Transmission time = Message size / data transmission rate = 1 x 106 x 8/ 200 x 103 = 40s 3-2008 UP-Copyrights reserved 17 6. What is the total delay (latency) for a message of size 5 million bits that is being sent on a link? The length of the link is 2000Km. The speed of electromagnetic wave inside the link is 2 × 108 m/s and the data transmission rate is 5Mbps? Solution Message size = 5 million bits = 5 x 106 Distance = 2000Km = 2 x 106 m Speed = 2 × 108 m/s data transmission rate = 5 × 106 b/s Latency = Propagation time + Transmission time 3-2008 UP-Copyrights reserved 18 Propagation time = Distance / Propagation speed = 2 x 106 /2 × 108 =1/100 = 0.01s Transmission time = Message size / data transmission rate = 5 x 106 / 5 x 106 = 1 s Latency = Propagation time + Transmission time = 0.01 3-2008 +1 UP-Copyrights reserved = 1.01s 19 8. Shown in the next figure, two signals with the same amplitude and phase, but different frequencies. Compare between the calculated periods for each signal. Solution Figure (a) : T = 1/F F =12 T = 1/12 = 0.083 s = 83 ms Figure (b) : T = 1/F F=6 T = 1/6 = 0.166 s = 166 ms 3-2008 UP-Copyrights reserved 20 9. A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have amplitude of 0. Draw the frequency domain of the signal. Solution Bandwidth = 200 kHz = Fh - Fl Where fh is the highest frequency, and fl is the lowest frequency. Then Fh – Fl = 200 (Fh+Fl)/2 = 140 The lowest frequency must be at 40 kHz and the highest at 240 kHz. 3-2008 UP-Copyrights reserved 21 20 V 3-2008 UP-Copyrights reserved 22 Unit Seconds (s) Equivalent 1s Unit hertz (Hz) Equivalent 1 Hz Milliseconds (ms) 10–3 s kilohertz (KHz) 103 Hz Microseconds (ms) 10–6 s megahertz (MHz) 106 Hz Nanoseconds (ns) 10–9 s gigahertz (GHz) 109 Hz Picoseconds (ps) 10–12 s terahertz (THz) 1012 Hz 3-2008 UP-Copyrights reserved 23